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Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q2 = +1.00 µC, and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a) r = 4.00 m, (b) r =0.700 m, and (c) r = 0.200 m? With V = 0 at infinity, what is V at (d) r = 4.00 m, (e) r = 1.00 m, (f) r = 0.700 m, (g) r = 0.500 m, (h) r = 0.200 m, and (i) r = 0? (j) Plot the E(r) and V(r) dependencies.

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okpalawalter8

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =[tex]1.685*10^3 N/C[/tex]

b E =[tex]36.69*10^3 N/C[/tex]

c E = 0 N/C

d [tex]V = 6.7*10^3 V[/tex]

e   [tex]V = 26.79*10^3V[/tex]

f   V = [tex]34.67 *10^3 V[/tex]

g   [tex]V= 44.95*10^3 V[/tex]

h    [tex]V= 44.95*10^3 V[/tex]

i    [tex]V= 44.95*10^3 V[/tex]

Explanation:

From the question we are given that

       The first charge [tex]q_1 = 2.00 \mu C = 2.00*10^{-6} C[/tex]

       The second charge [tex]q_2 =1.00 \muC = 1.00*10^{-6}[/tex]

      The first radius [tex]R_1 = 0.500m[/tex]

      The second radius [tex]R_2 = 1.00m[/tex]

 [tex]Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}[/tex]

And [tex]Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ][/tex]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           [tex]E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}[/tex]

                [tex]= 1.685*10^3 N/C[/tex]

Considering b

           [tex]r = 0.700 m \ , R_2 > r > R_1[/tex]

This implies that the electric field would be

            [tex]E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}[/tex]

             This because it the electric filed of the charge which is below it in distance that it would feel

            [tex]E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}[/tex]

               = [tex]36.69*10^3 N/C[/tex]

   Considering c

                      r  = 0.200 m

=>   [tex]r<R_1<R_2[/tex]

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> [tex]r > R_1 >r>R_2[/tex]

Now the potential difference is

                  [tex]V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V[/tex]

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m [tex]R_2 = r > R_1[/tex]

                [tex]V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V[/tex]

          Considering f

              [tex]r = 0.700 m \ , R_2 > r > R_1[/tex]

                      [tex]V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V[/tex]

          Considering g

             [tex]r =0.500\m , R_1 >r =R_1[/tex]

   [tex]V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V[/tex]

          Considering h

                [tex]r =0.200\m , R_1 >R_1>r[/tex]

  [tex]V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V[/tex]

           Considering i    

   [tex]r =0\ m \ , R_1 >R_1>r[/tex]

  [tex]V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V[/tex]

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